$\lim_{x \to \infty}e^{x-x^2}$ I am teaching calculus and am wanting to put the following question on a test
$$
\lim\limits_{x\rightarrow\infty}e^{x-x^2}
$$
I know this limit is 0, but as I try to show it rigorously (I mean justifying the steps, but no $\epsilon,\delta$) I am running into problems.
I know that $e^{x-x^2}=\frac{e^x}{e^{x^2}}$ but this doesn't simplify. Intuitively, $e^{x^2}$ grows faster than $e^x$ so the quotient approaches 0, but this seems a bit hand-wavey.
Maybe since $e^x$ is continuous, we can apply the limit directly to $x-x^2$ and get nonsense like $\infty-\infty^2$. So if we factor we get $x(1-x)$ and note that $x\rightarrow\infty$ while $1-x\rightarrow-\infty$ so we get $\infty\cdot-\infty$ which again intuitively is $-\infty$ but it doesn't really make sense to multiply infinities. Nonetheless, if we "accept" this and abuse the notation we get $\frac{1}{e^{\infty}}$ which goes to 0 as well.
I feel like I am over complicating this and I don't want to ask a question that I can't explain clearly. How would you explain this without using an $\epsilon,\delta$ proof?
EDIT: Thanks for all of the responses. Obviously there are multiple ways to think about this, but with how I have been teaching using $e^{x-x^2}=e^{x^2(\frac{1}{x}-1)}$ which goes to $\frac{1}{e^{x^2}}$ as $x\rightarrow\infty$ seems like the "best" way to explain it. I do however like the squeeze theorem argument and might make it a bonus to prove this limit using squeeeze.
 A: Note that since $e^x>x$
$$0\le e^{x-x^2}=\frac{1}{e^{x^2-x}}\le\frac{1}{x^2-x}\to0$$
thus for squeeze theorem
$$\lim\limits_{x\rightarrow\infty}e^{x-x^2}=0$$
A: hint
Since
$$\lim_{x\to+\infty}x (1-x)=-\infty ,$$
$$(\forall \epsilon>0) \;\; (\exists A>0 ) \;:$$
$$x>A \implies x (1-x)<\ln (\epsilon) $$
 $$\implies e^{x-x^2}<\epsilon $$
then
$$\lim_{x\to+\infty}e^{x-x^2}=0$$
A: Hint:
1)use continuity of exponential function.
2) use $e^{x-x^2}=e^{x^2(\frac{1}{x}-1)}$
3) note that $\lim_{x \to \infty}(\frac{1}{x}-1)=-1$ 
A: if you accept that $\lim_{x\to\infty}e^x = \infty$ then certainly also:
$$
\lim_{x\to\infty}\frac{e^x}e = \infty
$$
now for $x\gt1$ we have 
$$
\bigg(\frac{e^x}e \bigg)^x \gt \frac{e^x}e
$$
so
$$
\lim_{x\to\infty}\bigg(\frac{e^x}e \bigg)^x = \infty
$$
i.e.
$$
\lim_{x\to\infty} e^{x^2-x} = \infty
$$
and
$$
\lim_{x\to\infty} e^{x-x^2} = 0
$$
A: Hint: you have for large $x$ :$ \exp(- x^2)\leq\exp(x - x^2)\leq \frac{1}{x²}$ then use squeez theorem you get $\lim \exp(x - x^2)=0$ for $x \to +\infty$
A: $x - x^2 \sim - x^2$ in $\infty$ so $x-x^2 \to -\infty$ in $\infty$ hence $\exp(x - x^2) \to 0$ in $\infty$
A: Another attempt:
Let $x\gt 0:$
$0\lt e^{-x^2} e^x\lt e^{-x^2} e^{x^2/2} =$
$e^{-x^2/2}  = \dfrac{1}{e^{x^2/2} }$,  for $x \gt 2.$
Can you take it from here ?
A: There is a simple solution using your simplification:
We rise x to the power:
$$e^{x-x^2}=\frac{e^x}{e^{x^2}}=(\frac{e}{e^{2}})^x$$
and we know that $$\frac{e}{e^{2}}<1$$
So we conclude that:
$$\lim\limits_{x\rightarrow\infty}e^{x-x^2} = \lim\limits_{x\rightarrow\infty}(\frac{e}{e^{2}})^x = 0$$
